Question

Construct the angles of the following measurements:

$\left(i\right){30}^0\left(ii\right){75}^0\left(iii\right){105}^0\left(iv\right){135}^0\left(v\right){15}^0\left(vi\right)22{\frac{1}{2}}^{\circ }$

Answer 1

(i) ${30}^{\circ }$

Steps of construction :

(a) Draw a line segment BC.

(b) With centre B and a suitable radius draw an arc meeting BC at E.

(c) Cut off arc EF and bisect it at G. So that $\angle ABC={30}^{\circ }$.

(ii) ${75}^{\circ }$

Steps of Construction :

(a) Draw a line segment BC.

(b) With centre B and a suitable radius, draw an arc meeting BC at E.

(c) Cut off arc EF = FG from E.

(d) Bisect the arc FG at K and join BK so that $\angle KBC={90}^{\circ }$.

(e) Now bisect arc HF at L and join BL and produce it to A so that $\angle ABC={75}^{\circ }$.

(iii) ${105}^{\circ }$

Steps of construction :

(a) Draw a line segment BC.

(b) With centre B and a suitable radius draw an arc meeting BC at E.

(c) From E, cut off arc EF = FG and divide FG at H.

(d) Join BH meeting the arc at K.

(e) Now bisect the arc KG at L.

Join BL and produce it to A.

Then $\angle ABC={105}^{\circ }$.

(iv) ${135}^{\circ }$

Steps of construction :

(a) Draw a line segment BC.

(b) With centre B and a suitable radius, draw an arc meeting BC at E.

(c) From E, cut off EF = FG = GH.

(d) Bisect arc FG at K, and join them.

(e) Bisect arc KH at L.

(f) Join BL and produce it to A, then $\angle ABC={135}^{\circ }$.

(v) ${15}^{\circ }$

Steps of construction :

(a) Draw a line segment BC.

(b) With centre B and a suitable radius, draw an arc meeting BC at E.

(c) Cut off arc EF from E and bisect it at G.

Then $\angle GBC={30}^{\circ }$.

(d) Again bisect the arc EJ at H.

(e) Join BH and produce it to A.

Then $\angle ABC={15}^{\circ }$.

(vi) $22{\frac{1}{2}}^{\circ }$

Steps of construction:

(a) Draw a line segment BC.

(b) With centre B and a suitable radius, draw an arc and from E, cut off EF = FG.

(c) Bisect FG at H so that $\angle HBC={90}^{\circ }$.

(d) Now bisect $\angle HBC$ at K. So that $\angle YBC={45}^{\circ }$

(e) Again bisect $\angle YBC$ at J. So that $\angle ABC=22{\frac{1}{2}}^{\circ }.$